Why Define the Box Topology in Terms of a Redundant Basis? Let $\{X_i\}$ be an arbitrary collection of topological spaces.
Let $X = \prod X_i$.
I have seen in Munkres' textbook on Topology, for example, that we can define the box topology $\tau$ on $X$ via the basis $\mathcal{B} = \{\prod U_i : U_i$ is open in $X_i\}$.
But why define $\tau$ in terms of what $\mathcal{B}$ generates if they are the same sets (am I mistaken that they are indeed the same sets)?  Then my question specifically is if, under this definition, $\tau = \mathcal{B}$, why discuss a basis at all?
 A: Yes, you are mistaken; for example, consider the box topology on $\mathbb{R}\times\mathbb{R}$ (which is actually just the same as the normal topology). Let $U=(0,1)\subset\mathbb{R}$, which is an open set. Then
$$(U\times\mathbb{R})\cup (\mathbb{R}\times U)$$
is a "thick plus sign" in $\mathbb{R}^2$, which cannot be expressed as $V\times W$ for any open $V,W\subset \mathbb{R}$.
A: In general $\tau\supsetneqq\mathscr{B}$. For a simple example, consider $X=\square^\omega\Bbb R$, the box product of countably infinitely many copies of $\Bbb R$. Let $z$ be the point whose coordinates are all $0$. Then $X\setminus\{z\}$ is an open set not in $\mathscr{B}$.
For that matter, let $X$ be any Hausdorff space with at least two points. Let $\Delta=\{\langle x,x\rangle:x\in X\}$, the diagonal in $X\times X$ (which has the same topology both as a Tikhonov and as a box product). Then $X\setminus\Delta$ is an open set not in $\mathscr{B}$. To see this, let $x,y\in X$ with $x\ne y$; clearly $\langle x,y\rangle,\langle y,x\rangle\notin\Delta$. However, if $\langle x,y\rangle,\langle y,x\rangle\in U\times V$, then $x,y\in U\cap V$, and therefore $\langle x,x\rangle,\langle y,y\rangle\in(U\times V)\cap\Delta$.
